Let $G$ be an algebra of sets and define $\mathcal{M}(G)$ to be the smallest monotone class containing $G$. Then $\mathcal{M}(G)$ is precisely the $\sigma$-algebra generated by $G$, i.e. $\sigma(G) = \mathcal{M}(G)$.

On proofwiki.org, I've found a proof of a modified version of the statement where $G$ doesn't need to be an algebra, but only closed under complement. I can't find any mistake in their proof.

Since any other textbook contains the Wikipedia version of the statement, I thought being an algebra would be a crucial assumption.

So, is there any mistake in the proofwiki.org proof that I didn't see?

1 Answer
1

The proofwiki claim and proof looks correct, but I don't think this is as big of a deal as you seem to think. Clearly if $G$ is an algebra or a collection of subsets closed under complements, then $M(G) \subseteq \sigma(G)$. The bulk of the monotone class theorem is that $M(G)$ is large enough to be all of $\sigma(G)$. So if we only start off with $G$ being a collection of subsets closed under complements, then $M(G)$ in particular will contain all finite unions of sets in $G$ and so will already contain the algebra generated by $G$. A more concise relevant statement is $M(M(G)) = M(G)$, so since we just care about this being large enough, we might as well pretend $G$ is an algebra to start off with.